Determine resolution of use by modelling it as single slit

In summary, you are to find the width of a slit illuminated by a single source. You need to account for the fact that there are two sources, separated by an angle of 0.00015 radians.
  • #1
twinklestar28
21
0

Homework Statement



Model the pupil of the eye using a single slit. If the eye is able to notice changes in intensity of 20% show that it can distinguish two objects separated by an angle of 0.00015 radians if =500nm and diameter=4mm.



Homework Equations


I=Iosin^2((π/λ)asinθ)/ ((π/λ)asinθ)^2
sin θ=λ/a


The Attempt at a Solution



I attempted to find a (width) from the intensity equation but it doesn't seem to work, so I am unsure of how to start the calculation, the question is worth 4 marks, any help would be appreciated :)
 
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  • #2
twinklestar28 said:

Homework Statement



Model the pupil of the eye using a single slit. If the eye is able to notice changes in intensity of 20% show that it can distinguish two objects separated by an angle of 0.00015 radians if =500nm and diameter=4mm.

Homework Equations


I=Iosin^2((π/λ)asinθ)/ ((π/λ)asinθ)^2
sin θ=λ/a

The Attempt at a Solution



I attempted to find a (width) from the intensity equation but it doesn't seem to work, so I am unsure of how to start the calculation, the question is worth 4 marks, any help would be appreciated :)
Looks like for the slit width you are to use the diameter of the pupil ("Model the pupil of the eye using a single slit")

You wrote I(θ) for a slit illuminated by a single source. You need to account for the fact that there are two sources, separated by an angle of 0.00015 radians.
 
  • #3
O ok so the equation would be

I=Iosin^2((π/λ)asinθ-∅)/ ((π/λ)asinθ-∅)^2

it gives a hint to consider the intensity halfway between the peaks of the two images so i did

((π/λ)asin(θ-∅)) = ∏/2

but where do i put the intensity change of 0.2% to get (θ-∅) = 0.00015?
 
  • #4
twinklestar28 said:
O ok so the equation would be

I=Iosin^2((π/λ)asinθ-∅)/ ((π/λ)asinθ-∅)^2
That's the intensity of a single source that is located an angle ∅ away from the central axis.

But we have two sources. Let's say one of them is an angle ∅ away from the central axis, and one of them is on the central axis. You need to consider the intensity due to each source, and combine them to get the intensity in the combined diffraction pattern.
 
  • #5
Hello, I am fairly new here. I posted one new question but I forgot how to do that. Can anybody assist me by telling me how to post a new question?
 

What is the single slit model and how is it used to determine resolution?

The single slit model is a mathematical representation of the behavior of light as it passes through a narrow opening. It is used to determine the resolution of a system, which is the ability to distinguish between two closely spaced objects or points.

How does the width of the slit affect the resolution in the single slit model?

The width of the slit directly affects the resolution in the single slit model. As the slit width decreases, the resolution increases, allowing for the differentiation of smaller details. Conversely, a wider slit will result in lower resolution and the inability to distinguish between closely spaced objects.

What is the relationship between the wavelength of light and the resolution in the single slit model?

The resolution in the single slit model is inversely proportional to the wavelength of light. This means that as the wavelength increases, the resolution decreases, and vice versa. This is due to the diffraction of light as it passes through the slit, causing interference patterns that affect the resolution.

How does the distance between the slit and the screen affect the resolution in the single slit model?

The distance between the slit and the screen also affects the resolution in the single slit model. As the distance increases, the resolution decreases, as the light spreads out and the interference patterns become less distinct. This is why using a magnifying lens can improve the resolution in certain systems.

Can the single slit model be applied to other systems besides light?

Yes, the principles of the single slit model can be applied to other systems besides light, such as sound waves or water waves. In these cases, the width of the slit would represent the size of the opening or obstacle that the wave is passing through, and the wavelength would correspond to the frequency of the wave.

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